Appendix: All Solutions to the Problems
873
The boundary conditions are y(0) = y(l) = 0, y
(0) = y
(l) = 0.
7.11 Let the definite problem
2 u = f (x, y)(x, y ∈ D)
u| Γ = 0,
∂u
∂n
Γ = 0
where,
2 is the biharmonic operator; Γ is the boundary of the domain D. Prove
that
2 is a symmetric positive definite operator and ((
2 u, u) = ((u, ,u), and find
out the variational problem equivalent to the definite problem.
Solution It has been proved from problem 1.5 that the biharmonic operator
2 is a
positive definite operator, namely
v
2 u = uv + ∇ · (v∇w) − ∇ · (w∇v)
where, w = u.
Making the inner product ((
2 u, v), using the Green formula and the boundary
conditions u| Γ = 0,
∂u
∂n
Γ = 0, there is
((
2 u, v) =
¨
D
v
2 udD =
¨
D
[vu + ∇ · (v∇w) − ∇ · (w∇v)]dD
=
¨
D
vudD +
Γ
v
∂w
∂n
− w
∂v
∂n
dΓ
=
¨
D
vudD +
Γ
v
∂u
∂n
− u
∂v
∂n
dΓ =
¨
D
vudD
Transposing v and u, the above result remains unchanged, so
2 is a symmetric
operator.
Putting v = u, then
((
2 u, u) =
¨
D
uudD = ((u, ,u)
When u = 0, there is
((
2 u, u) =
¨
D
uudD =
¨
D
((u)
2 dD > 0
This shows that
2 is also a positive definite operator. Thus, the functional
equivalent to the differential equation can be established
J [u] = ((
2 u, u) − 2u f =
¨
D
[((u)
2
− 2u f ]dD